Proving the Limit of (2x^2 + y^2)/(x^2 + y^2) as (x,y) --> (-1,2)

[Haftred]

I am trying to prove the limit of the function (2x^2 + y^2) / (x^2 + y^2) as (x,y) ---> (-1 , 2) is 6/5.

So I have 0 < sq((x+1)^2 + (y-2)^2) < delta

and f(x) - 6/5 < Epsilon

I found a common denominator and made epsilon the quotient of two polynomials. Also, i recognized that you could factor the numerator to yield some function of delta. However, the denominator is becoming a problem. I get that the numerator is less than 4D^2 + 8D + D^2 + 4D, after using the triangle inequality, but i don't know what to do with the denominator. Am i close to the right method, or am i totally doing it wrong? Is proving this even possible?

 

[Edgardo]

I'm not sure, but couldn't you use something similar to the limit theorems for functions, as in
example 4 here: http://www.tpub.com/math2/37.htm ?

This may be helpful:
http://www.math.oregonstate.edu/hom...usQuestStudyGuides/vcalc/limcont/limcont.html

 

[murshid_islam]

how about converting into polar coordinates?

 

[HallsofIvy]

The question is, what are you allowed to use? It is obvious that both numerator and denominator are continuous functions and the denominator does not go to 0. By the "limit theorems" it obviously goes to 6/6 since the numerator goes to 6 and the denominator goes to 5. Are you saying that you are required to do an epsilon-delta proof?

 

[Haftred]

yes, it's obvious what the limit is, but I need to prove it using the delta-epsilon proof.

 

[HallsofIvy]

Then your delta will measure distance from (-1, 2). It might be best to use murshid islam's suggestion: convert to polar coordinates. But you will also need to translate (-1, 2) to the origin. That is, x= -1+ r cos($\theta$), y= 2+ r sin($\theta$). That way, $\delta= r$.

 

[Haftred]

Ok converting to polar coordinates seems like a good idea; however, the denominator is still causing me problems.

I will show all the work I have done so far:

We want to show that the limit of the function:

$$\frac{2x^2 + y^2}{x^2+ y^2}$$ is 1.2 as one approaches the point (-1,2).

We want to use the $$\delta - \epsilon$$ proof:

$$x = -1 + r\cos\Theta$$
$$y = 2 + r\sin\Theta$$

$$r < \delta$$

$$\epsilon < f(x) - 1.2 = \frac{4x^2 - y^2}{5x^2+5^2}$$

I substituted the values of x and y in terms of $$r$$ and $$\Theta$$

And I end up with:

ab((4r^2cos^(t) - r^2sin^2(t) - 8rcos(t) - 4rsin(t) / (5r^2 - 10rcos(t) + 20rsin(t) + 25)) < Epsilon

I sill cannot prove it if I substitute delta for 'r'; the denominator is stil causing problems.

 

[Haftred]

sorry x = -1 + rcos(t) is what i used, not 1 + rcos(t)